Exercise 5.4.2

Exercise statement

Prove the remaining claims in Proposition 5.4.7.

Proposition 5.4.7. All the claims in Proposition 4.2.9 which held for rationals, continue to hold for real numbers.

Let $x,y,z$ be real numbers. Then the following properties hold.

(a) (Order trichotomy) Exactly one of the three statements $x=y$ , $x < y$ , or $x > y$ is true.
(b) (Order is anti-symmetric) One has $x < y$ if and only if $y > x$ .
(c) (Order is transitive) If $x < y$ and $y < z$ , then $x < z$ .
(d) (Addition preserves order) If $x < y$ , then $x+z < y+z$ .
(e) (Positive multiplication preserves order) If $x < y$ and $z$ is positive, then $xz < yz$ .

Hints

None.

How to think about the exercise

This is a straightforward exercise, so I don’t have anything to say.

Model solution

(a) First we show that at least one of the possibilities is true. Consider the number $x-y$ . By Proposition 5.4.4, it is zero, positive, or negative. If $x-y$ is zero, then $x=y$ . If $x-y$ is positive, then $x > y$ by definition of order. If $x-y$ is negative, then $x < y$ by definition of order. Thus in each case, at least one of the possibilities is true.

Now we show that at most one of the possibilities is true. If $x=y$ and $x > y$ , then $x-y$ is both zero and positive, which contradicts Proposition 5.4.4. Similarly if $x=y$ and $x < y$ , then $x-y$ is both zero and negative, a contradiction. Also if $x > y$ and $x < y$ , then $x-y$ is both positive and negative, which is again a contradiction.

(b) Suppose $x < y$ . Then $x-y$ is negative. Thus by Proposition 5.4.4, we see that $-(x-y) = y-x$ is positive. This means $y > x$ , which is what we wanted to show. Conversely, suppose $y > x$ . Then $y-x$ is positive, so by Proposition 5.4.4 we see that $-(y-x) = x-y$ is negative, which means $x < y$ as required.

(c) Suppose $x < y$ and $y < z$ . By part (b), this means $y > x$ and $z > y$ , so $y-x$ and $z-y$ are positive. By Proposition 5.4.4, we thus see that $(y-x) + (z-y) = z-x$ is a positive number, so that $z > x$ . By part (b) again, this means $x < z$ as required.